If `"sin" (pi "cos" theta) = "cos" (pi "sin" theta),` then which one fo the following is correct?

Options

A. `"cos" theta = (3)/(sqrt(3))`
B. `"cos" (theta - (pi)/(2)) = (1)/(sqrt(2))`
C. `"cos" (theta - (pi)/(4)) = (1)/(2sqrt(2))`
D. `"cos" (theta + (pi)/(4)) = (1)/(2sqrt(2))`

Answer

To solve the equation \( \sin(\pi \cos \theta) = \cos(\pi \sin \theta) \), we will follow these steps:

### Step 1: Use the co-function identity
We know that \( \cos(90^\circ - x) = \sin(x) \). We can rewrite the equation using this identity:
\[ \sin(\pi \cos \theta) = \cos(\pi \sin \theta) \implies \sin(\pi \cos \theta) = \sin\left(\frac{\pi}{2} - \pi \sin \theta\right) \]

### Step 2: Set the arguments equal
Since the sine function is periodic, we can set the arguments equal to each other:
\[ \pi \cos \theta = \frac{\pi}{2} - \pi \sin \theta \]

### Step 3: Simplify the equation
Rearranging gives us:
\[ \pi \cos \theta + \pi \sin \theta = \frac{\pi}{2} \]
Dividing through by \( \pi \) (assuming \( \pi \neq 0 \)):
\[ \cos \theta + \sin \theta = \frac{1}{2} \]

### Step 4: Use the identity for sine and cosine
We can rewrite \( \cos \theta + \sin \theta \) using the identity:
\[ \cos \theta + \sin \theta = \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) \]
Thus, we have:
\[ \sqrt{2} \sin\left(\theta + \frac{\pi}{4}\right) = \frac{1}{2} \]

### Step 5: Isolate the sine function
Dividing both sides by \( \sqrt{2} \):
\[ \sin\left(\theta + \frac{\pi}{4}\right) = \frac{1}{2\sqrt{2}} \]

### Step 6: Solve for the angle
We know that \( \frac{1}{2\sqrt{2}} = \sin\left(\frac{\pi}{4}\right) \). Therefore, we can write:
\[ \theta + \frac{\pi}{4} = n\pi + \frac{\pi}{4} \quad \text{or} \quad \theta + \frac{\pi}{4} = n\pi + \frac{3\pi}{4} \]
for \( n \in \mathbb{Z} \).

### Step 7: Solve for \( \theta \)
From the first equation:
\[ \theta = n\pi \]
From the second equation:
\[ \theta = n\pi + \frac{\pi}{2} \]

### Conclusion
The solutions for \( \theta \) are:
\[ \theta = n\pi \quad \text{or} \quad \theta = n\pi + \frac{\pi}{2} \]